 ## Presentation on theme: "Quadratic Functions."— Presentation transcript:

A Parabola.

A quadratic function can be expressed in three formats: f(x) = ax2 + bx + c is called the standard form is called the factored form f (x) = a(x - h)2 + k, is called the vertex form

The Standard Form f(x)=ax2 + bx + c If a > 0 then parabola opens up
If a < 0 then parabola opens down (0,c) is the y-intercept Axis of symmetry is

The Vertex Form f (x) = a(x - h)2 + k, a  0
***y – k = a(x-h)2 The graph of f is a parabola whose vertex is the point (h, k) The parabola is symmetric to the line x = h If a > 0, the parabola opens upward if a < 0, the parabola opens downward

Graphing Parabolas-Vertex Form
To graph f (x) = a(x - h)2 + k: Determine whether the parabola opens up or down. If a > 0, it opens up. If a < 0, it opens down. Determine the vertex of the parabola--- V(h, k). Find any x-intercepts. Replace f (x) with 0. Solve the resulting quadratic equation for x. Find the y-intercept by replacing x with zero. Plot the intercepts and vertex. Connect these points with a smooth curve that is shaped like a cup.

The Factored Form Factored Form of a quadratic
r represents roots or x intercepts Axis of symmetry:

Vertex form f (x) = a(x - h)2 + k
Example Graph the quadratic function f (x) = -2(x - 3)2 + 8. Vertex form f (x) = a(x - h)2 + k   a=-2 h =3 k = 8 Given equation f (x) = -2(x - 3)2 + 8 This is a parabola that opens down with vertex V(3,8)

Example cont. The x-intercepts are 1 and 5.
Find the vertex. The vertex of the parabola is at (h, k). Because h = 3 and k = 8, the parabola has its vertex at (3, 8). Find the x-intercepts. Replace f (x) with 0: f (x) = -2(x - 3)2 + 8. 0 = -2(x - 3) Find x-intercepts, setting f (x) equal to zero. 2(x - 3)2 = 8 (x - 3)2 = Divide both sides by 2. (x - 3) = ± Apply the square root method. x - 3 = or x - 3 = Express as two separate equations. x = or x = Add 3 to both sides in each equation. The x-intercepts are 1 and 5. The parabola passes through (1, 0) and (5, 0).

Example cont. Graph the parabola. With a vertex at (3, 8)
f (0) = -2(0 - 3) = -2(-3) = -2(9) + 8 = -10 Find the y-intercept. Replace x with 0 in f (x) = -2(x - 3)2 + 8. The y-intercept is –10. The parabola passes through (0, -10). Graph the parabola. With a vertex at (3, 8) x-intercepts at (1,0) and (5,0) and a y-intercept at (0,–10) the axis of symmetry is the vertical line x = 3.

The Vertex of a Parabola Whose Equation Is f (x) = ax 2 + bx + c
Consider the parabola defined by the quadratic function f (x) = ax 2 + bx + c. The parabola's vertex is at

Example Graph the quadratic function f (x) = -x2 + 6x -.
Determine how the parabola opens. Note that a, the coefficient of x 2, is -1. Thus, a < 0; This negative value tells us that the parabola opens downward. Find the vertex. x = -b/(2a). Identify a, b, and c to substitute the values into the equation for the x-coordinate: x = -b/(2a) = -6/2(-1)=3. The x-coordinate of the vertex is 3. We substitute 3 for x in the equation of the function to find the y-coordinate: y=f(3) = -(3)^2+6(3)-2= =7, the parabola has its vertex at (3,7).

Example Graph the quadratic function f (x) = -x2 + 6x -.
Find the x-intercepts. Replace f (x) with 0 in f (x) = -x2 + 6x - 2. 0 = -x2 + 6x (If you cannot factor)

Example Graph the quadratic function f (x) = -x2 + 6x -.
Find the y-intercept. Replace x with 0 in f (x) = -x2 + 6x - 2. f (0) = • = - The y-intercept is –2. The parabola passes through (0, -2). Graph the parabola.

Consider f(x) = ax2 + bx +c. If a > 0, then f has a minimum that occurs at x = -b/(2a). This minimum value is f(-b/(2a)). If a < 0, the f has a maximum that occurs at x = -b/(2a). This maximum value is f(-b/(2a)).

The Discriminant and the Kinds of Solutions to ax2 + bx +c = 0
No x-intercepts No real solution; two complex imaginary solutions b2 – 4ac < 0 One x-intercept One real solution (a repeated solution) b2 – 4ac = 0 Two x-intercepts Two unequal real solutions b2 – 4ac > 0 Graph of y = ax2 + bx + c Kinds of solutions to ax2 + bx + c = 0 Discriminant b2 – 4ac